Question

R. Find the number of a and B- particles emitted in the process 222Rn 214 po 86 84​

Answer 1

Let there be x α and y β particles be emitted.

92

238

→

82

214

Pb+x

2

4

He+y

−1

0

e

Equating atomic numbers and atomic masses we get,

92=82+2x−y

⟹2x−y=10

238=214+4x

⟹x=6

y=2

So 6 alpha particles and 2 beta particles.